Question

A bulb emits light of wavelength 4500 Angstrom. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second.

Answer 1

Each photon has energy of  hν  
h = planks constant
ν = frequency of light
If there are N photons emitted then the total energy radiated is  Nhν .
Average power is given by  P=Energy/time=E/t=Nhν/t
But  N/t=n  is the number of photons emitted per second.
Therefore,  n=P/hν .
ν=c/λ
So,  n=Pλ/hc
If all the power is not radiated then the effective power = rated power x efficiency.
So the number of photons emitted per second  n=Peffλ/c = P×ηλ/hc  where η is the efficiency.
In the problem power P = 150W, wavelength λ = 4500A and efficiency is 8% = 0.08
So number of photons emitted per second, 
n=(150×0.08*4500*10-10) /(6.6×10−34) *(3*×108) = 27.2*1018 
27.2*1018 number of photons will be emitted per second by the bulb
Hope this will help you
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Answer 2

Answer : $2.72\times 10^{19}$ photons are emitted by the bulb per second.

Explanation:

Energy of one photon = $E=\frac{hc}{\lambda }$

$E=\frac{6.62\times 10^{-34} J/s\times 3\times 10^{8}m/s}{4500\times 10^{-10} m}=4.41\times 10^{-19} J$

$E_n=n\times 4.41\times 10^{-19} J$ = Total energy of n numbers of photons

The bulb is rated as 150 watt and 8% of the energy is emitted as light:

Power = 150 watts

Time =  1 s

$Energy=Power\times time= 150 watts\times 1 s = 150 J$

Total energy of bulb = 150 J.

Energy emitted as light=$=E_L=150 J\times \frac{8}{100} J$

$E_n=E_L$

$E_n=n\times 4.41\times 10^{-19} J=150 J\times \frac{8}{100} J$

$n=2.72\times 10^{19} photons$

$2.72\times 10^{19}$ photons are emitted by the bulb per second.